题目链接：点击打开链接
题意：
给定一个n层书架，一共取m本书。
下面n行给出每层书的价值。
每次可以取任意一层的最左端或最右端的一本书。
问能获得的最大价值。
思路：
1、显然是先求出对于每层任取任意本书能获得的最大价值。
2、然后背包一下。
1：
对于一层书任意j本，那么一定是从左端取k本，右端取 j-k本，求个前缀和然后枚举 j和k即可。每层n^2的dp
2：
分组背包。
import java.io.BufferedReader;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import java.text.DecimalFormat;import java.util.ArrayDeque;import java.util.ArrayList;import java.util.Arrays;import java.util.Collection;import java.util.Collections;import java.util.Comparator;import java.util.Deque;import java.util.HashMap;import java.util.Iterator;import java.util.LinkedList;import java.util.Map;import java.util.PriorityQueue;import java.util.Scanner;import java.util.Stack;import java.util.StringTokenizer;import java.util.TreeMap;import java.util.TreeSet;import java.util.Queue;import java.io.File;import java.io.FileInputStream;import java.io.FileNotFoundException;import java.io.FileOutputStream;public class Main {int m, n;int[] sum = new int[N], num = new int[N];int[][] dp = new int[N][N], h = new int[2][M];void work() throws Exception{n = Int(); m = Int();for(int i = 1; i <= n; i++){num[i] = Int();sum[0] = 0;for(int j = 1; j <= num[i]; j++){ sum[j] = sum[j-1]+Int(); }dp[i][0] = 0;for(int j = 1; j <= num[i]; j++){dp[i][j] = 0;for(int k = 0; k <= j; k++)dp[i][j] = max(dp[i][j], sum[k] + sum[num[i]] – sum[num[i]-j+k]);}}int cur = 0, old = 1;for(int i = 0; i <= m; i++)h[cur][i] = 0;for(int i = 1; i <= n; i++){cur ^= 1; old ^= 1;for(int j = 0; j <= m; j++)h[cur][j] = h[old][j];for(int j = 0; j <= num[i]; j++){for(int k = j; k <= m; k++)h[cur][k] = max(h[cur][k], h[old][k-j]+dp[i][j]);}}out.println(h[cur][m]);}public static void main(String[] args) throws Exception{Main wo = new Main();in = new BufferedReader(new InputStreamReader(System.in));out = new PrintWriter(System.out); // in = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt")))); // out = new PrintWriter(new File("output.txt"));wo.work();out.close();}static int N = 101;static int M = 10001;DecimalFormat df=new DecimalFormat("0.0000");static int inf = (int)1e9;static long inf64 = (long) 1e18;static double eps = 1e-8;static double Pi = Math.PI;static int mod = (int)1e9 + 7 ;private String Next() throws Exception{while (str == null || !str.hasMoreElements())str = new StringTokenizer(in.readLine());return str.nextToken();}private int Int() throws Exception{return Integer.parseInt(Next());}private long Long() throws Exception{return Long.parseLong(Next());}private double Double() throws Exception{return Double.parseDouble(Next());}StringTokenizer str;static Scanner cin = new Scanner(System.in);static BufferedReader in;static PrintWriter out; /* class Edge{int from, to, dis, nex;Edge(){}Edge(int from, int to, int dis, int nex){this.from = from;this.to = to;this.dis = dis;this.nex = nex;}}Edge[] edge = new Edge[M<<1];int[] head = new int[N];int edgenum;void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}void add(int u, int v, int dis){edge[edgenum] = new Edge(u, v, dis, head[u]);head[u] = edgenum++;}/**/int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;int pos = r;r–;while (l <= r) {int mid = (l + r) >> 1;if (A[mid] <= val) {l = mid + 1;} else {pos = mid;r = mid – 1;}}return pos;}int Pow(int x, int y) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}double Pow(double x, int y) {double ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}int Pow_Mod(int x, int y, int mod) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}long Pow(long x, long y) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}long Pow_Mod(long x, long y, long mod) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}int gcd(int x, int y){if(x>y){int tmp = x; x = y; y = tmp;}while(x>0){y %= x;int tmp = x; x = y; y = tmp;}return y;}int max(int x, int y) {return x > y ? x : y;}int min(int x, int y) {return x < y ? x : y;}double max(double x, double y) {return x > y ? x : y;}double min(double x, double y) {return x < y ? x : y;}long max(long x, long y) {return x > y ? x : y;}long min(long x, long y) {return x < y ? x : y;}int abs(int x) {return x > 0 ? x : -x;}double abs(double x) {return x > 0 ? x : -x;}long abs(long x) {return x > 0 ? x : -x;}boolean zero(double x) {return abs(x) < eps;}double sin(double x){return Math.sin(x);}double cos(double x){return Math.cos(x);}double tan(double x){return Math.tan(x);}double sqrt(double x){return Math.sqrt(x);}}
，未曾失败的人恐怕也未曾成功过。
